Question:

A vessel contains equal volumes of $ \text{S}{{\text{O}}_{\text{2}}} $ and $ \text{C}{{\text{H}}_{4}}. $ Through a small hole the gases effused into vacuum. After 200 seconds the total volume is reduced to half. What is the ratio of $ \text{S}{{\text{O}}_{\text{2}}} $ and $ \text{C}{{\text{H}}_{\text{4}}} $ remaining in the vessel?

Updated On: Aug 15, 2024
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The Correct Option is B

Solution and Explanation

According to Grahams law of diffusion, $ \frac{{{r}_{S{{O}_{2}}}}}{{{r}_{C{{H}_{4}}}}}=\sqrt{\frac{{{M}_{C{{H}_{4}}}}}{{{M}_{S{{O}_{2}}}}}} $ $ \frac{{{r}_{S{{O}_{2}}}}}{{{r}_{C{{H}_{4}}}}}=\sqrt{\frac{16}{64}}=\sqrt{\frac{1}{4}}=\frac{1}{2} $ $ {{r}_{S{{O}_{2}}}}:{{r}_{C{{H}_{4}}}}=1:2 $ $ \text{S}{{\text{O}}_{\text{2}}} $ and $ \text{C}{{\text{H}}_{4}} $ effused in the ratio 1 : 2. So, the amount remains in the vessel will be in the ratio as $ S{{O}_{2}}:C{{H}_{4}}=2:1 $
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